Selections from Question One

a) 1.69 J / (4.184 J/cal) =

b) 0.3587 J / (4.184 J/cal) =

c) 820.1 J / (1000 J/kJ) = 0.8201 kJ

d) 68 cal x (4.184 J/cal) =

e) 423 cal / (1000 cal/kcal) - 0.423 kcal

f) 20.0 cal x (4.184 J/cal) =

m) 9.806 kJ x (1000 J/kJ) = 9806 J

o) 5467.9 kcal --> 5467900 cal x (4.184 J/cal) =

2b. q = (20.0 g) (22.4 °C) (4.184 J g¯^{1} °C¯^{1})

2d. q = (25.0 g) (25.0 °C) (2.06 J g¯^{1} °C¯^{1})

3. q = (3.21 g) (4.0 °C) (4.184 J g¯^{1} °C¯^{1})

4. q = (55.6 g) (34.9 °C) (4.184 J g¯^{1} °C¯^{1}). How did I know this was liquid water?

5. Same type as #4, but no answer. Good luck!!

6b. (6.02 kJ/mol) (74.5 g / 18.0 g/mol)

6d. (40.7 kJ/mol) (43.89 g / 18.0 g/mol)

7. Some hints:

a) water means liquid water8c. (2.29)(48.9 - x)(4.184) = (3.65)(x-36.1)(4.184); 243.746 = 5.94x; x = 41.0 °C

b) assume the first four have no temp. change. They then become one step problems.

c) question 2e is a two part solution, 2f is three parts as is 2g.

d) 2f and 2g are same 3 steps just in reverse order.

9. A student places 42.3 grams of ice at 0.0 °C in an insulated bottle. The student adds 255.8 grams of water at 90.0 °C. Determine the final temperature of the mixture.

**Solution**

The solution to this problem is complicated by the fact that ice is added to hot water. Heat will flow from the hot water to the ice in order to melt it. HOWEVER (and this is important), when the ice melts, it does not change in temperature.

So what we need to do first is figure out how much energy is needed to melt the ice and then figure out what happens to the 90 degree water's temperature. Here goes:

**1) Figure the energy to melt ice**

q = (ΔH_{fus}) (moles of ice)q = (6.02 kJ/mol) (42.3 g / (18.0 g/mol)) = 14.147 kJ

I'll ignore significant figures until the final answer.

Now, this energy comes out of the hot water, so let's see how much a temperature drop this involves.

**2) Temperature drop by 90 degree water**

q = (mass) (Δt) (C_{p})14,147 J = (255.8 g) (x) (4.184 J/g °C) = 13.22 °C

Notice that I have changed the kJ to J to agree with the specific heat value I am using.

Since this is a Δt, I 90 minus 13.22 to get 76.78 °C as the final temperature of the water. However, we are not done. All we have done is create a new problem and here it is:

What is the final temperature when 42.3 g of water at 0.0 °C and 255.8 grams of water at 76.78 °C are mixed?

The technique covered in worksheet #1 is used and the answer the ChemTeam gets is 65.9 °C.

10. (6020 J/mol)(21.4 g / 18.0 g/mol) + (21.4 g)(x - 0)(4.184 J/g °C) = (40700 J/mol)(13.1 g / 18.0 g/mol) + (13.1 g)(0 - x)(4.184 J/g °C)

The above is (ice melts) + (liquid temp up) = (steam condenses) + (liquid temp down).

16. (450.2) (95.2 - x) (0.089) = (60.0) (x - 10) (4.184)

17. (x) (85.0 - 12.5) (0.045) = (6020) (54.0 / 18.0) + (54.0) (12.5) (4.184)

Notice that the unknown is mass of iron. That means that every other part of the problem MUST be a number. You can't have two variables and only one equation,if you want a numerical solution.

18. Determine the final temperature when 45.8 grams of aluminum at -5.2 °C is added to a mixture of 45.0 grams of ice at 0.0 °C and 2000.0 grams of water at 95.0 °C.

**Solution**

This problem is like the one above, except that it has one more step due to the aluminum. Here is what to do:

1) Figure out the energy needed to melt the ice. (15.05 kJ)

2) Figure out how much the 95 degree water drops in temperature. (1.8 °C)

3) Figure out the equilibrium temperature between the 45 grams (of melted ice) and the 2000 grams (at the new temperature, lower than 95) (91.2 °C)

4) (Here's the new step.) Figure out the equilibrium temperature between the aluminum and the 2045.0 grams of water at whatever temp. Notice: 2045, NOT 2000.

For the last step, with proper sig figs, I get 91.2, which is essentially the same answer as step #3. This is because of the small specific heat of the aluminum (0.089 J/g °C).

19. A sample of cobalt, A, with a mass of 5.00 g, is initially at 25.0 °C. When this sample gains 6.70 J of heat, the temperature rises to 27.9 °C. Another sample of cobalt, B, with a mass of 6.50 g, is initially at 25.0 °C. If sample B gains 5.00 J of heat, what is the final temperature of sample B. (Hint: think about the specific heat of both samples.)

a) calculate the specific heat of sample A:

6.70 J = (5.00 g) (2.9 °C) (x); x = 0.462 J / g °C

b) use the specific heat to calculate Δt for sample B

5.00 J = (6.40 g) (x) (0.462 J / g °C)Δt = 1.7 °C, final temp. = 26.7 °C

20. 50.0 g of copper at 200.0 °C is placed in ice at 0 °C. How many grams of ice will melt?

a) determine heat lost by the Cu as it cools from 20.0 down to zero:

q = (50.0 g) (200.0 °) (0.39 J/g °C) = 3900 J

b) calculate mass of ice melted by 3900 J:

3900 J = (x) (334.16 J/g) = 11.7 g(334.16 J/g is the heat of fusion for water in Joules per gram)